Exam MAS-I — Poisson Processes Flashcards

1. **$N(0)=0$** — the count starts at zero. 2. **Independent increments** — counts over disjoint time intervals are independent. 3. **Stationary increments** — the distribution of $N(t+s)-N(s)$ depends only on the length $t$, not on $s$, and equals $\text{Poisson}(\lambda t)$. Equivalently, in a small interval of length $h$ the probability of one event is $\lambda h + o(h)$, of two or more is $o(h)$, and these are independent across disjoint intervals.

**Stationary:** the count in any interval depends only on its length — $N(t+s)-N(s)\sim\text{Poisson}(\lambda t)$ for every $s$. **Independent:** counts in non-overlapping intervals are independent random variables. Together they imply the process "forgets" its past: the number of future arrivals after time $s$ is independent of the history up to $s$ and behaves like a fresh Poisson process started at $0$. This is the process-level analogue of the exponential interarrival's **memoryless** property.

For a small width $h$: $\Pr(N(t+h)-N(t)=1)=\lambda h + o(h)$, $\Pr(N(t+h)-N(t)\ge 2)=o(h)$, $\Pr(N(t+h)-N(t)=0)=1-\lambda h + o(h)$, with increments over disjoint intervals independent. Here $o(h)$ denotes a term with $\lim_{h\to0}o(h)/h=0$. This local description is equivalent to the global axioms and is the basis of the Poisson differential equations.

$N(t)\sim\text{Poisson}(\lambda t)$, so $\Pr(N(t)=k)=e^{-\lambda t}\frac{(\lambda t)^{k}}{k!}$ for $k=0,1,2,\dots$. Mean $E[N(t)]=\lambda t$ and variance $\text{Var}(N(t))=\lambda t$ — for any Poisson variable the mean equals the variance. The rate $\lambda$ is the expected number of events per unit time.

$N(t+s)-N(s)\sim\text{Poisson}(\lambda t)$ — it depends only on the interval length $t$. The starting point $s$ drops out by **stationary increments**, and by **independent increments** this increment is independent of $N(s)$ (the count before time $s$). So you may always shift any interval to start at $0$ when computing increment probabilities.

With $t=1$ day, $N(1)\sim\text{Poisson}(\lambda t)=\text{Poisson}(3)$. $\Pr(N(1)=2)=e^{-3}\frac{3^{2}}{2!}=e^{-3}\frac{9}{2}=4.5\,e^{-3}$. Since $e^{-3}\approx 0.049787$, $\Pr(N(1)=2)\approx 4.5(0.049787)\approx 0.2240$.

Take $t=0.5$ hour, so $\lambda t = 4(0.5)=2$ and $N(0.5)\sim\text{Poisson}(2)$. $\Pr(N(0.5)\ge 1)=1-\Pr(N(0.5)=0)=1-e^{-2}\frac{2^{0}}{0!}=1-e^{-2}$. Since $e^{-2}\approx 0.135335$, the probability $\approx 1-0.135335 = 0.8647$.

Over $t=3$ weeks the count is $N(3)\sim\text{Poisson}(\lambda t)=\text{Poisson}(6)$. $\Pr(N(3)=5)=e^{-6}\frac{6^{5}}{5!}=e^{-6}\frac{7776}{120}=64.8\,e^{-6}$. Since $e^{-6}\approx 0.00247875$, $\Pr(N(3)=5)\approx 64.8(0.00247875)\approx 0.1606$.

The two hours are disjoint intervals, so by **independent increments** the counts are independent, each $\sim\text{Poisson}(1.5)$. $\Pr(N_1=3)=e^{-1.5}\frac{1.5^{3}}{3!}=e^{-1.5}\frac{3.375}{6}=0.5625\,e^{-1.5}\approx 0.5625(0.223130)\approx 0.12551$. $\Pr(N_2=1)=e^{-1.5}(1.5)\approx 1.5(0.223130)\approx 0.33470$. Product $\approx 0.12551 \times 0.33470 \approx 0.0420$.

By independent increments, $N(5)-N(3)$ is independent of $N(3)$, and we need $N(5)-N(3)=8-3=5$ events in the interval $(3,5]$ of length $2$. By stationary increments that increment is $\text{Poisson}(\lambda\cdot 2)$. For $\lambda=2$: $N(5)-N(3)\sim\text{Poisson}(4)$, so $\Pr=e^{-4}\frac{4^{5}}{5!}=e^{-4}\frac{1024}{120}\approx 8.5333(0.018316)\approx 0.1563$.

The interarrival times $T_1,T_2,\dots$ (gaps between consecutive events) are i.i.d. $\text{Exponential}(\lambda)$ with density $f(t)=\lambda e^{-\lambda t}$ for $t\ge 0$. Each has mean $E[T_i]=\frac{1}{\lambda}$ and variance $\frac{1}{\lambda^{2}}$. The exponential interarrival is equivalent to the Poisson-counting description — either fully characterizes the process.

If $T\sim\text{Exponential}(\lambda)$, then $\Pr(T>s+t \mid T>s)=\Pr(T>t)=e^{-\lambda t}$ for all $s,t\ge0$. Having already waited $s$ with no event gives no information about the remaining wait — the residual time is again $\text{Exponential}(\lambda)$. The exponential is the only continuous distribution with this property; it is why the Poisson process "restarts" at every instant.

$S_n=T_1+\cdots+T_n$ is the sum of $n$ i.i.d. $\text{Exponential}(\lambda)$ gaps, so $S_n\sim\text{Gamma}(n,\lambda)$ (an Erlang) with density $f_{S_n}(t)=\frac{\lambda^{n}t^{n-1}e^{-\lambda t}}{(n-1)!}$ for $t\ge0$. Mean $E[S_n]=\frac{n}{\lambda}$ and variance $\text{Var}(S_n)=\frac{n}{\lambda^{2}}$.

The $n$-th arrival occurs after time $t$ **iff** fewer than $n$ events have happened by time $t$: $\{S_n>t\}=\{N(t)<n\}=\{N(t)\le n-1\}$. Hence $\Pr(S_n>t)=\sum_{k=0}^{n-1}e^{-\lambda t}\frac{(\lambda t)^{k}}{k!}$, which is exactly the survival function of a $\text{Gamma}(n,\lambda)$. This identity converts gamma-tail questions into Poisson sums.

The wait to the next event is $\text{Exponential}(\lambda)$ with $\lambda=6$ per hour. Expected wait $=\frac{1}{\lambda}=\frac{1}{6}$ hour $=10$ minutes. For more than $20$ minutes $=\frac{1}{3}$ hour: $\Pr(T>\tfrac{1}{3})=e^{-\lambda t}=e^{-6/3}=e^{-2}\approx 0.1353$.

The $4$th-arrival time is $S_4\sim\text{Gamma}(4,2)$. $E[S_4]=\frac{n}{\lambda}=\frac{4}{2}=2$ days. $\text{Var}(S_4)=\frac{n}{\lambda^{2}}=\frac{4}{4}=1$, so the standard deviation is $\sqrt{1}=1$ day.

Use $\Pr(S_3>1)=\Pr(N(1)<3)=\Pr(N(1)\le 2)$ with $N(1)\sim\text{Poisson}(3)$. $\Pr(N(1)=0)=e^{-3}=0.049787$, $\Pr(N(1)=1)=3e^{-3}=0.149361$, $\Pr(N(1)=2)=4.5e^{-3}=0.224042$. Sum $\approx 0.049787+0.149361+0.224042=0.4232$.

By the **memoryless** property, conditioning on no event in the first $2$ minutes leaves the residual waiting time again $\text{Exponential}(\lambda=1)$. Expected additional wait $=\frac{1}{\lambda}=1$ minute — the same as the unconditional mean. The two minutes already elapsed are irrelevant.

Given $N(t)=n$, the $n$ unordered arrival times are distributed as $n$ independent $\text{Uniform}(0,t)$ random variables; the ordered arrival times $S_1<S_2<\cdots<S_n$ are distributed as the **order statistics** of those $n$ uniforms. The joint density of the ordered times is $\frac{n!}{t^{n}}$ on $0<s_1<\cdots<s_n<t$. Crucially this is **free of $\lambda$** — the rate affects how many events occur, not where they land given the count.

Since each of the $n$ arrivals is independently $\text{Uniform}(0,t)$, each lands in $(0,s]$ with probability $p=\frac{s}{t}$ independently. So the count in $(0,s]$ given $N(t)=n$ is $\text{Binomial}\!\left(n,\frac{s}{t}\right)$. Its conditional mean is $n\cdot\frac{s}{t}$. This is the standard way conditioning-on-the-total problems are solved.

With $n=1$, the single arrival time is $\text{Uniform}(0,t)$ — equally likely anywhere in the interval. Its expected value is $\frac{t}{2}$ (the midpoint), independent of $\lambda$. For example, if exactly one claim occurred in a $10$-hour shift, its expected time of occurrence is $5$ hours in.

Given $N(2)=3$, each event is $\text{Uniform}(0,2)$, landing in $(0,1]$ with $p=\frac{1}{2}$. The count in the first hour is $\text{Binomial}(3,\tfrac{1}{2})$. $\Pr(2\text{ in first hour})=\binom{3}{2}\left(\tfrac{1}{2}\right)^{2}\left(\tfrac{1}{2}\right)^{1}=3\cdot\frac{1}{8}=\frac{3}{8}=0.375$. Note the rate $\lambda=4$ never enters — only the count and the interval split matter.

Conditioned on $N(10)=5$, the $5$ arrival times are i.i.d. $\text{Uniform}(0,10)$, each with mean $\frac{10}{2}=5$. Expected sum $=5\times 5 = 25$. More generally, given $N(t)=n$ the expected sum of arrival times is $n\cdot\frac{t}{2}$.

Given $N(2)=4$, each arrival is $\text{Uniform}(0,2)$; the chance one falls in the second hour $(1,2]$ is $p=\frac{1}{2}$. The count in the second hour is $\text{Binomial}(4,\tfrac{1}{2})$, so $\Pr(\text{all }4)=\left(\tfrac{1}{2}\right)^{4}=\frac{1}{16}=0.0625$.

If each event of a rate-$\lambda$ Poisson process is independently classified as type-1 with probability $p$ (and type-2 with probability $1-p$), then the type-1 events form a Poisson process with rate $\lambda p$ and the type-2 events form a Poisson process with rate $\lambda(1-p)$, and these two processes are **independent** of each other. With $k$ categories of probabilities $p_1,\dots,p_k$ (summing to $1$), you get $k$ independent Poisson processes with rates $\lambda p_1,\dots,\lambda p_k$.

If $N_1(t),\dots,N_m(t)$ are **independent** Poisson processes with rates $\lambda_1,\dots,\lambda_m$, then their sum $N(t)=N_1(t)+\cdots+N_m(t)$ is a Poisson process with rate $\lambda=\lambda_1+\cdots+\lambda_m$. Given an event of the merged process, it came from stream $j$ with probability $\frac{\lambda_j}{\lambda}$, independently of everything else — the dual of thinning.

By thinning, large claims form a Poisson process with rate $\lambda p = 10(0.30)=3$ per day, independent of the small-claim process. Large claims in a day: $\text{Poisson}(3)$, expected $3$. Over $5$ days: $\text{Poisson}(3\times 5)=\text{Poisson}(15)$, expected $15$.

By thinning, trucks form a Poisson process with rate $\lambda p = 20(0.15)=3$ per minute, so the truck count in a minute is $\text{Poisson}(3)$. $\Pr(2\text{ trucks})=e^{-3}\frac{3^{2}}{2!}=4.5\,e^{-3}\approx 4.5(0.049787)\approx 0.2240$.

By superposition the combined process is Poisson with rate $\lambda=\lambda_1+\lambda_2=4+6=10$ per hour, so the total in one hour is $\text{Poisson}(10)$. $\Pr(3\text{ total})=e^{-10}\frac{10^{3}}{3!}=e^{-10}\frac{1000}{6}\approx 166.667(0.0000453999)\approx 0.00757$.

By superposition the merged rate is $\lambda=3+2=5$. Each event is independently attributed to stream $j$ with probability $\frac{\lambda_j}{\lambda}$. $\Pr(\text{from stream 1})=\frac{\lambda_1}{\lambda_1+\lambda_2}=\frac{3}{5}=0.6$.

Buyers are a thinned Poisson process with rate $\lambda p = 12(0.25)=3$ per hour. Over $t=0.5$ hour the buyer count is $\text{Poisson}(3\times 0.5)=\text{Poisson}(1.5)$. $\Pr(4\text{ buyers})=e^{-1.5}\frac{1.5^{4}}{4!}=e^{-1.5}\frac{5.0625}{24}=0.210938\,e^{-1.5}\approx 0.210938(0.223130)\approx 0.04707$.

A nonhomogeneous Poisson process has a time-varying rate $\lambda(t)\ge0$. It keeps independent increments but **drops** stationary increments. The **mean function** is $m(t)=\int_0^{t}\lambda(u)\,du$, and the count over $(a,b]$ is $\text{Poisson}\!\left(\int_a^{b}\lambda(u)\,du\right)=\text{Poisson}(m(b)-m(a))$. In particular $N(t)\sim\text{Poisson}(m(t))$ with $E[N(t)]=m(t)$.

Compute the expected count over the interval, $\mu=\int_a^{b}\lambda(u)\,du = m(b)-m(a)$, then use the Poisson law: $\Pr(N(b)-N(a)=k)=e^{-\mu}\frac{\mu^{k}}{k!}$. Unlike the homogeneous case, the interval's location matters because $\lambda(t)$ varies — only the integrated intensity $\mu$ enters the formula.

$E[N(3)]=m(3)=\int_0^{3}2t\,dt=\left[t^{2}\right]_0^{3}=9$. So $N(3)\sim\text{Poisson}(9)$ with expected count $9$ events.

The expected count over $(1,2]$ is $\mu=\int_1^{2}2t\,dt=\left[t^{2}\right]_1^{2}=4-1=3$, so the increment is $\text{Poisson}(3)$. $\Pr(2\text{ events})=e^{-3}\frac{3^{2}}{2!}=4.5\,e^{-3}\approx 4.5(0.049787)\approx 0.2240$.

$\mu=m(2)=\int_0^{2}(3+u)\,du=\left[3u+\tfrac{u^{2}}{2}\right]_0^{2}=6+2=8$, so $N(2)\sim\text{Poisson}(8)$. $\Pr(N(2)=0)=e^{-8}\frac{8^{0}}{0!}=e^{-8}\approx 0.000335$.

$X(t)=\sum_{i=1}^{N(t)}Y_i$, where $N(t)$ is a Poisson process with rate $\lambda$ and the $Y_i$ are i.i.d. (the per-event sizes, e.g. claim amounts), independent of $N(t)$. By the compound-mean formula, $E[X(t)]=E[N(t)]\,E[Y]=\lambda t\,E[Y]$ — the expected number of events times the expected size.

Using the compound-variance formula $\text{Var}(X)=E[N]\text{Var}(Y)+\text{Var}(N)E[Y]^{2}$ with $E[N(t)]=\text{Var}(N(t))=\lambda t$: $\text{Var}(X(t))=\lambda t\,\text{Var}(Y)+\lambda t\,E[Y]^{2}=\lambda t\,E[Y^{2}]$. The two terms collapse because the Poisson count has equal mean and variance, leaving the clean result $\text{Var}(X(t))=\lambda t\,E[Y^{2}]$.

$X(1)=\sum_{i=1}^{N(1)}Y_i$ is compound Poisson with $\lambda t = 5(1)=5$. $E[X(1)]=\lambda t\,E[Y]=5(200)=\$1{,}000$. The expected daily aggregate claim is $\$1{,}000$.

First $E[Y^{2}]=\text{Var}(Y)+E[Y]^{2}=150^{2}+200^{2}=22500+40000=62500$. $\text{Var}(X(1))=\lambda t\,E[Y^{2}]=5(62500)=312500$. Standard deviation $=\sqrt{312500}\approx \$559.02$.

For $Y\sim\text{Uniform}(0,10)$: $E[Y]=5$ and $E[Y^{2}]=\frac{10^{2}}{3}=\frac{100}{3}\approx 33.333$ (since $E[Y^2]=\text{Var}(Y)+E[Y]^2=\frac{100}{12}+25$). With $\lambda t = 3(2)=6$: $E[X(2)]=\lambda t\,E[Y]=6(5)=30$. $\text{Var}(X(2))=\lambda t\,E[Y^{2}]=6\left(\frac{100}{3}\right)=200$.

$E[X]=\lambda t\,E[Y]=100(500)=50000$. $\text{Var}(X)=\lambda t\,E[Y^{2}]=100(400000)=40{,}000{,}000$, so $\text{SD}(X)=\sqrt{4\times10^{7}}\approx 6324.56$. Standardize: $z=\frac{60000-50000}{6324.56}\approx 1.5811$. $\Pr(X>60000)\approx 1-\Phi(1.5811)\approx 1-0.9431=0.0569$.